The solubility of Pb`(OH)_2` in pure water in pure water is `20/3xx10^(-6)M`. Calculate concentration of `OH^(-)` ion in a buffer solution of `P^(H)=8`.

Options

A. `20/3xx10^(-6)M`
B. `10/3xx10^(-6)M`
C. `10^(-6)M`
D. `2xx10^(-6)M`

Answer

To solve the problem, we need to calculate the concentration of hydroxide ions (OH⁻) in a buffer solution with a given pH of 8.

### Step-by-Step Solution:

1. **Understand the relationship between pH and H⁺ concentration:**
The pH of a solution is defined as:
\[ \text{pH} = -\log[\text{H}^+] \]
Given that the pH is 8, we can find the concentration of hydrogen ions (H⁺).

2. **Calculate the concentration of H⁺ ions:**
Rearranging the equation for pH gives us:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-8} \]
Therefore, the concentration of H⁺ ions is:
\[ [\text{H}^+] = 10^{-8} \, \text{M} \]

3. **Use the ion product of water to find OH⁻ concentration:**
The product of the concentrations of H⁺ and OH⁻ ions in water at 25°C is given by:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
We can rearrange this to find the concentration of OH⁻ ions:
\[ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{10^{-8}} \]

4. **Calculate OH⁻ concentration:**
Substituting the values:
\[ [\text{OH}^-] = 10^{-14} \times 10^{8} = 10^{-6} \, \text{M} \]

5. **Final Result:**
The concentration of hydroxide ions (OH⁻) in the buffer solution at pH 8 is:
\[ [\text{OH}^-] = 10^{-6} \, \text{M} \]

### Summary of the Solution:
The concentration of OH⁻ ions in a buffer solution with pH 8 is \( 10^{-6} \, \text{M} \).